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I have proven that the Cantor set is uncountable using Cantors diagonalization argument in base $3$. Now I have to prove that it has cardinality $\mathfrak{c}$ due to the fact that it is uncountable but I am struggling to even get started.

Any help would be appreciated.

dahaka5
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  • I have looked at this but I need to deduce that the cardinality is $c$ because it is uncountable – dahaka5 Nov 27 '17 at 14:51
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    By writing numbers in base 3 you can give a bijection with thr space of sequences having value 0 or 1. – vap Nov 27 '17 at 14:51
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    Since $C \subset [0,1]$, so cardinality of $C$ is atmost $\mathfrak{c}$, since it is uncountable, cardinality of $C$ is actually $\mathfrak{c}$ by continuum hypothesis – Chinnapparaj R Nov 27 '17 at 14:58
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    @ChinnapparajR That is a very very bad answer. The continuous hypothesis is not true in standard mathematics. – Stella Biderman Nov 27 '17 at 14:59
  • @StellaBiderman I think, rather, the CH is undecidable in "standard mathematics" (ZF+ choice)). – David Mitra Nov 27 '17 at 15:07
  • @DavidMitra There are models of ZFC in which CH is true and there are models of ZFC in which CH is false. It is neither true nor false, and so it is "not true". – Stella Biderman Nov 27 '17 at 15:18
  • @StellaBiderman, I am not a mathematician, so I'm interested to know if your use of "not true" is conventional. If a mathematician says that something is "not true in ZFC", does that mean that it could be "independent of ZFC" or "undecidable in ZFC"? Or more generally, when mathematicians say that something is not true (in any context), does that not imply that it is false? Anytime that I read in a math text that something is "not true", should I keep in mind that it might not be false either? – Joe Mar 19 '21 at 14:42
  • @Joe I have a typo of sorts in that comment. I had intended to write There are models of ZFC in which CH is true and there are models of ZFC in which CH is false. It is neither true nor false, and so it is not "true". (note the change in where the quotes are). I would say that it is conventional in 99% of mathematics to use "not true" and "false" interchangeably. We are specifically having a conversation about logical foundations though, which makes things a bit different. – Stella Biderman Mar 19 '21 at 15:45
  • Ok, thank you. I have only vague familiarity with logic, and don’t if the “law of the excluded middle” is rigorous, or if it applies to the foundations of maths. – Joe Mar 19 '21 at 16:16

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If you’re trying to reason solely based off the fact that the cantor set is uncountable (this is my impression from your comments), then you cannot prove that it has cardinality $\mathfrak{c}$ with the standard axioms of mathematics. You need some form of the continuum hypothesis. As mentioned in Andre, if you can use the fact that the cantor set is closed, then you can prove it has carnality $\mathfrak{c}$ because every perfect subset of $\mathbb{R}$ has cardinality $\mathfrak{c}$

You can also do slightly more sophisticated proofs based on the algebraic representation of the cantor set. If you can consider the function that “reinterprets” a base 3 expression as a base 2 expression (sending $0$ to $0$ and $2$ to $1$). This function is a surjection into $\mathbb R$ and so the Cantor set is no smaller than $\mathbb R$. Since it’s also a subset, we get equality between their sizes.

Stella Biderman
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  • Your first paragraph probably needs some massaging: Any uncountable closed subset of the reals contains a perfect subset, and any perfect subset of $\mathbb R$ has size $\mathfrak c$. It may well be this is the argument that was expected. – Andrés E. Caicedo Nov 27 '17 at 17:57